Infinite groups in which all irreducible representations are one-dimensional. For a finite group $G$, if all of its irreducible representations are one-dimensional, then by complete reducibility, we know that $G$ is an abelian group. So what about infinite group? Is there an infinite group, non-abelian, but all of its irreducible representations are one-dimensional?
 A: If you restrict attention to finite-dimensional representations, then such groups exist; that is, there are infinite nonabelian groups all of whose finite-dimensional irreducible representations are $1$-dimensional. In fact there are infinite nonabelian groups all of whose finite-dimensional representations are trivial; see this math.SE answer for an argument that a finitely presented nonabelian simple group, such as the Higman group, has this property. Other examples exist such as simple groups of cardinality larger than the continuum, e.g. $PSL_2(K)$ where $K$ is a  field of cardinality strictly larger than the continuum, but that construction feels more awkward and unnatural to me (when was the last time you ran into a field of cardinality strictly larger than the continuum?).
For infinite-dimensional irreducible representations the answer may depend delicately on exactly what you consider a representation (e.g. is it on a topological vector space, if so what kind of topology) and what you consider an irreducible representation (e.g. do you want no invariant subspaces or no closed invariant subspaces).
A simple option is to ask for Hilbert space representations which are irreducible in the sense that there are no closed invariant subspaces; in that case any group $G$ has a Hilbert space representation on $\ell^2(G)$. I am not very familiar with how this representation behaves for infinite discrete groups but I believe it is always a direct integral of irreducible representations (in the Hilbert space sense); if that's true, then if $G$ is nonabelian there must be some irreducible representation with nonabelian image and hence which cannot be $1$-dimensional (or else the direct integral couldn't be faithful). Hopefully someone who actually knows how this stuff works can comment.
